Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds." If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}^{-}\). Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?

Step by step solution

01

Recall the definition of electron affinity

Electron affinity is the energy change when an electron is added to a neutral atom in the gaseous state, forming a negative ion.

02

Compare the electron affinity of hydrogen with alkali elements and halogens

We need to compare the electron affinity of hydrogen with alkali elements and halogens. Hydrogen has an electron affinity value of around -73 kJ/mol. Alkali elements have similar values, whereas halogens have much higher values (more negative). This is because halogens have one fewer electron than the next noble gas configuration and are hence more eager to gain an additional electron to achieve a stable electron configuration.

03

Analyze the similarity in electron affinity between hydrogen and alkali elements

Hydrogen, like alkali elements, has only one electron in its valence shell, and they all need one more electron to reach the electron configuration of the nearest noble gas. This similarity in their electron configurations is the reason why hydrogen's electron affinity is closer to alkali metals than to halogens. #Part (b)#

04

Analyze the statement about hydrogen's bonding atomic radius

The given statement says that "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds." We will now determine if this statement is true or not.

05

Compare the bonding atomic radius of hydrogen with other elements

Hydrogen's bonding atomic radius is around 31 pm, which is indeed the smallest among all the elements that form chemical compounds. The small size is due to its simple electron configuration, with only one electron in its 1s orbital.

06

Conclusion and explanation

The statement is true. Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds. The reason is due to its electron configuration (1s^1), with only one electron in the 1s orbital. #Part (c)#

07

Recall the definition of ionization energy

Ionization energy is the energy required to remove an electron from a gaseous atom or ion.

08

Compare the ionization energy of hydrogen with alkali elements and halogens

We need to compare the ionization energy of hydrogen with alkali elements and halogens. Hydrogen has an ionization energy value of 1312 kJ/mol, which is closer to the values of halogens than alkali metals.

09

Analyze the similarity in ionization energy between hydrogen and halogens.

Ionization energy is related to the electron configuration and effective nuclear charge. Hydrogen (1s^1) and halogens (ns^2 np^5) both have high ionization energies due to their relatively stable electron configurations. The full or nearly full valence shells result in a stronger effective nuclear charge, making it harder to remove an electron from hydrogen and halogens compared to alkali metals. #Part (d)#

10

Define the first ionization energy

The first ionization energy refers to the energy required to remove the first electron from a neutral atom in its gaseous state.

11

Write the process for the first ionization energy of the hydride ion

For the hydride ion, H^-, which has the electron configuration of 1s^2, the first ionization energy process is as follows: \(\mathrm{H^{-} \rightarrow H + e^{-}}\) #Part (e)#

12

Write the process for the electron affinity of a neutral hydrogen atom

The process for the electron affinity of a neutral hydrogen atom in its gaseous state is: \(\mathrm{H + e^{-} \rightarrow H^{-}}\)

13

Compare the processes in part (d) and (e)

On comparing the process for the first ionization energy of the hydride ion (H^− → H + e^−) with the process for the electron affinity of a neutral hydrogen atom (H + e^− → H^−), we can see that both processes are inverse of each other. The first ionization energy of the hydride ion represents the energy required to remove an electron from the ion, while the electron affinity of a neutral hydrogen atom represents the energy change when an electron is added to the neutral atom.

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